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A six person committee composed of Alice, Be, Connie, Dolph, Egbert, and Francisco is to select a chairperson, secretary, and treasurer.

a) How many selections exclude Connie?
b) How many selections are there in which neither Ben nor Francisco is an officer?
c) How many selections are there in which both Ben and Francisco are officers?
d) How many selections are there in which Dolph is an officer and Francisco is not an officer?
e) How many selections are there in which either Dolph is chairperson or he is not an officer?
f) How many selections are there in which Ben is either chairperson or treasurer?

My solutions:
a) $5 * 4 * 3$
b) $4 * 3 * 2$
c) $3 * 2 * 4$ ($3$ spots can be chosen for Ben, $2$ spots left to choose from for Francisco, and 4 members available for the last position)
d) $3 * 4 * 3$ ($3$ spots can be chosen for Dolph, 4 members left to choose from, then 3 members left to choose from)
e) $1 * 5 * 4$ ($1$ way to choose chairperson, $5$ members to choose next position, $4$ members left to choose from for last position)
f) $x = {}$set in which Ben is chairperson${} = 20$
$y ={}$set in which Ben is treasurer${} = 20$
$x + y = 40$

Marc van Leeuwen
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J. Doe
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  • Question unclear. Where does the officer "come from" all of the sudden????? – barak manos Mar 30 '16 at 05:46
  • @barakmanos an officer is someone who is in the leadership. Meaning they are chairperson, secretary, or treasurer. – J. Doe Mar 30 '16 at 05:48
  • So why didn't you write "How many selections are there in which Connie is not an officer?"??? What's the point in making the question ambiguous? – barak manos Mar 30 '16 at 05:52
  • @barakmanos I wrote it exactly as it is written on my assignment. They probably wrote it that way to be able to include the word "exclude" to give a hint about solving it. – J. Doe Mar 30 '16 at 05:57
  • E) is definitely incorrect. You only counted was Dolph can be chairman. Not was he is no officer. – fleablood Mar 30 '16 at 06:34
  • Otherwise very good – fleablood Mar 30 '16 at 06:35

3 Answers3

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a, b, c, and d are right. e is 80 ($1*5*4=20$ when Dolph is chairperson, $5*4*3=60$ when he is not an officer). For f, you should write $|X|=20$, $|Y|=20$, $|X\cup Y|=40$.

v7d8dpo4
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Part e will be the sum of $2$ cases:

Case 1 : Dolph is a chairperson

Number of ways =$1*5*4=20$

Case 2: Dolph isn't elected

Number of ways =$5*4*3=60$

Hence a total of $80$ ways for part $e$

(Rest all parts seem correct)

Nikunj
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a) How many selections are there in which Connie is not an officer?

$$\binom{6-1}{3}\cdot3!$$

b) How many selections are there in which neither Ben nor Francisco is an officer?

$$\binom{6-2}{3}\cdot3!$$

c) How many selections are there in which both Ben and Francisco are officers?

$$\binom{6-2}{3-2}\cdot3!$$

d) How many selections are there in which Dolph is an officer and Francisco is not an officer?

$$\binom{6-2}{3-1}\cdot3!$$

e) How many selections are there in which either Dolph is chairperson or he is not an officer?

$$\binom{6-1}{3-1}\cdot(3-1)!+\binom{6-1}{3}\cdot3!$$

f) How many selections are there in which Ben is either chairperson or treasurer?

$$\binom{6-1}{3-1}\cdot(3-1)!+\binom{6-1}{3-1}\cdot(3-1)!$$

barak manos
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